Algebra 1 mixture question

Question

A chemist combines 10% acid with 30% acid to make 500 milliliters of 15% acid. How much of each kind of acid is needed?

Answer 1

.3x + .1(500-x) = .15(500)

.3x -.1x = .15(500) -.1(500)

.2x = .05(500)

x = 25/.2 = 125 mL of 30% mixed with

500-125 = 375 mL of 10% yield

500 mL of 15%

10 is within 5 of 15

30 is within 15 of 15

15/5 = 3

3 times as much 10% is needed as 30%

375/125 = 3

Answer 2

Set up a set of linear equations where x = volume of the 10% acid, y = volume of 30% acid...

Eq 1) x + y = 500

Eq 2) 10x + 30y = 15*500

Note: in Equation 2, I'm ignoring the percent sign because it is in each term. Alternatively, you could write it as 0.1x + 0.3y = 0.15*500

And simplify...

Divide each term in equation 2 by 10 to get: x + 3y = 750

Rearrange equation 1 to get y in terms of x and you get: y = 500 - x

Put the new equation 1 y value into equation 2... x + 3(500-x) = 750 and solve

x + 1500 - 3x = 750

-2x + 1500 = 750

-2x = -750

x = 375 ml of the 10% solution

And from equation 1... y = 500 - x

y = 500-375 = 125 ml of the 30% solution